Idempotents which are not Murray-von Neumann equivalent to its adjoint

Question

What is an example of a $C^{*}$ algebra with an idempotent $e$ such that $e$ is not Murray-von Neumann equivalent to $e^{*}$?

Answer 1

An idempotent $e$ is always equivalent to $e^*$.

On the unitization of $A$, let $z=1+(e-e^*)^*(e-e^*)$. This is positive and invertible. By noticing that $z=1-e-e^*+ee^*+e^*e$, it is clear that $z$ commutes with $e$ and $e^*$, and a fortiori so does $z^{-1}$.

Let $x=ee^*\in A$, $y=e^*z^{-1}e\in A$ (recall that $A$ is an ideal in its unitization). Note that $$ze=ee^*e,\ \ \ ze^*=e^*ee^*;$$ then $$ xy=ee^*(e^*z^{-1}e)=z^{-1}ee^*e=e, $$ $$ yx=(e^*z^{-1}e)ee^*=z^{-1}e^*ee^*=e^*. $$

(I know the ideas behind this from Davidson, Proposition IV.1.1)

Edit: interestingly, a byproduct of this argument is that every idempotent is the product of two positives.